= Compound geometric distribution
{title2=$M_S(t)=p/(1-(1-p)M_X(t))$}
= Zero-based geometric compound sum
{synonym}
Take an independent <geometric distribution> count with $\mathbb P(N=k)=p(1-p)^k$, $k\ge0$, and sum its independent positive claims. Its <moment-generating function> is the displayed composition wherever finite. With $q=1-p$ and severity <moments> $m_j$, the third <cumulant> is $(q/p)m_3+3(q/p)^2m_1m_2+2(q/p)^3m_1^3$, which is strictly positive for nonzero positive claims and $0<p<1$. The law has an atom of mass $p$ at zero. Its positive-count convention would instead add a factor $M_X(t)$ to the numerator, as in the <geometric-sum moment-generating function>.
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